Problem: Rewrite the equation by completing the square. $x^{2}-16x+63 = 0$ $(x + $
Answer: Begin by moving the constant term to the right side of the equation. $x^2 - 16x = -63$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-16$, half of it would be $-8$, and squaring it gives us ${64}$. $x^2 - 16x { + 64} = -63 { + 64}$ We can now rewrite the left side of the equation as a squared term. $( x - 8 )^2 = 1$ This is equivalent to $(x+{-8})^2=1$